# LESSON 15 Chapter 14 Maneuvering Performance ANA Chapter 2 176-178 and Chapter 5

Chapter 14

Maneuvering Performance

*General Turning Performance

*In a turn, the airplane is not in a state of equilibrium, there must be a force to accelerate the turn.

*In a turn, not all the lift is opposite weight

*Effective lift or the vertical component of lift is opposite to weight

*Some lift is tilted into the horizontal component.

*General Turning Performance

* This requires the aircraft generate more total lift to the point at which effective lift will equal aircraft weight.

*The force that is generated toward the center of the turn is called centripetal force

*The horizontal component of lift is what generates this force

*The acceleration towards the center is called radial acceleration.

*General Turning Performance

*The force that is generated outward from the center of the turn is called centrifugal force

*Total lift then is offset by the G’s times the weight.

*Or G’s = lift divided by weight

*Load factor or G is defined as lift/weight or the combined force of centrifugal force and weight.

*Bank angle alone is the determining component for load factor.

*To solve for load factor for any bank angle we can use the formula:

*Φ=phi (pronounced phee)

*

*

*If we were to work this formula out to a 90 degree angle we would see that lift approaches infinity.

*So theoretically, flight with a 90 degree angle is impossible.

*The trick is the fuselage, tail and the vertical component of thrust make up the necessary lift requirement.

*Things that do not affect load factor are:

*Airspeed (as long as the aircraft does not stall)

*Aircraft type

*Power setting

*The equal and opposite reaction force to centripetal force is centrifugal force

*The plane’s inertia, is what causes this force

*So inertia opposes the horizontal component of lift

*Stall speed depends on the square root of the G loading or put another way, stall speed increases proportional to the square root of the load factor.

*So to figure stall speed at a specific bank angle, simply first compute the load factor for that particular bank, take the square root of the load factor then take that times the stall speed in unaccelerated level flight and presto, you’ve got the new stall speed at that bank angle.

*What?

*For example, if the plane stalls at 50 kts, at 60 degrees the load factor is 2.

*The square root of 2 is 1.4 so 1.4 x 50 = 70 kts.

*It is this principle which limits how we maneuver our airplanes.

*Consider what a 40 degree bank would do to a plane with a stall speed of 59 in the traffic pattern say turning from upwind to crosswind while doing a Vx climb.

*The LF at 40 degrees bank is about 1.3

*The root of the LF of 1.3 is 1.14

*So take 1.14x 59 = 67 kts

*Poof, stall speed is 2 kts higher than climb speed.

*Stall spin die.

*Thankyou, thankyou thankyou very much

*Intentionally stalling above Va tremendously effects LF

*If we rework the formula, we can figure the amount of G’s it will take to stall at a certain airspeed

*If an aircraft stalls at 50 kts, it would take 4 Gs to stall at 100kts

*If you could pull 9 Gs with this plane it would stall at 150kts

*So this means that the more airspeed the plane has, the higher the available load factor will be

*This also means that this is a function of the amount of dynamic pressure

*Remember dynamic pressure is IAS

*Wouldn’t it be cool if we had a graph that would indicate this?

*The Vg Diagram, Vn diagram, maneuver envelope or flight envelope

*The Vg diagram shows graphically how stall speed increases as the G’s go up.

*The curved portion shows the aerodynamic limits of the plane.

*The straight lines are basically the stall speed at the limit load factor.

*For the normal category airplane the load factors are +3.8 and -1.52

*Vne

*The Vg Diagram

*Normal Category +3.8, -1.52

*Utility Category    +4.4, -1.76

*Acrobatic Category +6.0, -3.0

*These structural limits are required to be imposed by FAR part 23 and assume:

*Gross weight

*Symmetrical loads (no rolling pull outs)

*No corrosion metal fatigue or other structural damage

*Smooth air no gusts

*A factor of 50% is assumed

*

*

*Vg Diagram

*The lines of maximum lift capability (curved lines) are the first items of importance on the Vg diagram

*The aircraft is capable of developing no more than +1 G at 62 mph, the wings level stall speed of the aircraft

*Since the maximum load factor varies with the square of the airspeed, the maximum positive lift capability of this aircraft is 2 G at 92 mph, 3 G at 112 mph, 4.4 G at 137 mph

*Any load factor above this line is unavailable aerodynamically

*The same situation exists for negative lift flight with the exception that the speed necessary to produce a given negative load factor is higher than that to produce the same positive load factor.

*If the aircraft is flown at a positive load factor greater than the positive limit load factor of 4.4, structural damage is possible.

*Vg Diagram

*There are two other points of importance on the Vg diagram.

*One point is the intersection of the positive limit load factor and the line of maximum positive lift capability (134kts)

*The airspeed at this point is the minimum airspeed at which the limit load can be developed aerodynamically

*Any airspeed greater than this provides a positive lift capability sufficient to damage the aircraft.

*Conversely, any airspeed less than this does not provide positive lift capability sufficient to cause damage from excessive flight loads

*The usual term given to this speed is maneuvering speed

*At Va minimum usable turn radius or maneuverability occurs at this condition

*The maneuver speed is a valuable reference point, since an aircraft operating below this point cannot produce a damaging positive flight load

*Any combination of maneuver and gust cannot create damage due to excess airload when the aircraft is below the maneuver speed

*

*Vg Diagram

*The other point of importance on the Vg diagram is the intersection of the negative limit load factor and line of maximum negative lift capability (100kts)

*Any airspeed greater than this provides a negative lift capability sufficient to damage the aircraft; any airspeed less than this does not provide negative lift capability sufficient to damage the aircraft from excessive flight loads

*The Vg Diagram

*Usually a separate gust penetration chart was provided by the manufacture in the past for certain airplanes for calculation of such.

*The stress imposed is a function of the lift to weight ratio.

*Remember that the amount of lift available to be generated varies with the square of the airspeed.

*Let’s say that a plane has a Va of 116kts.

*This means that if we were to increase the AOA rapidly to the stall AOA while doing 116kts our 2150lb plane would generate 8170lbs of lift before stalling at 3.8 g.

*This is the Va for the plane at 2150 but what about at other weights?

*Va must decrease with the square root of the Weight decrease.

*Lets say the same plane weighs 1500lbs and is again stalled at 116kts.

*The same amount of lift of 8170lbs is going to be generated by the wings as in the plane at gross weight.

*Since LF=lift/weight:

*The positive load factor now is 8170/1500 or 5.44 g.

*This is 1.64 Gs above the limit of 3.8

*So the argument goes, so what, the wings are still generating the same amount of lift of 8170lbs so no problem right?

*Wrong!

*The problem is with fixed weight components such as the engine, baggage, retractable landing gear etc.

*All these components are now acting on the fuselage at the rate of 5.44 Gs

*This places a hell of a stress on the spar carry through which is the piece that bolts your wing spars together

*Not only that but the mounts for these components, the baggage compartment floor ect are not stressed to handle that much weight

*Because of the lighter overall weight of the plane the fixed weight components are subjected to higher acceleration forces.

* The engine mounts may not be stressed to handle the weight of the engine at 5.4 gs.

* A baggage compartment that can hold 200lbs at 3.8 gs could keep it together with 760lbs of force but at 5.4 gs that same baggage weighs in at 1080lbs.

*Plus you could break your tail off

*Likewise if you put 400lbs in the baggage compartment and pulled 3.8 gs a total of 1520lbs is put on a baggage compartment floor stressed to handle 760lbs.

*This could be a good thing if it’s your mother in law stuffed in there

*And she’s wearing concrete overshoes

*And no parachute

*A really deep lake

*With piranha

*And sharks

*No let’s make that a Volcano

*Yea a Volcano

*An active Volcano

*With liquid hot magma

*

*The radius of turn goes up as you go faster at a given bank angle.

*Since we are dealing with how fast the plane is traveling through the air, this is a function of TAS

*In fact, the radius goes up as the V2 goes up.

*The radius of turn at 300kts is 4 times bigger as at 150kts.

*The formula is:

*Rate of Turn

*The rate of turn is based on bank angle.

* The higher the bank angle the quicker the rate of turn.

*The primary rate of turn we shoot for is 3 degrees/sec or 2min for a 360 turn also known as standard rate.

*Rate of Turn

*A easy formula for determining the standard rate bank angle is to take the airspeed divide by 10, half that number and add to the original.

*Or you can use Doles:

*

*There is a relationship between bank angle, velocity, rate and radius of turn. pgs. 236-237 shows that relationship

*The faster you go the tighter the turn until reaching Va

*Turns above Va may overstress the airframe because the LLF may be exceeded above this speed

*If operating above Va, the only option is to reduce bank angle to prevent exceeding the LLF

*At Va, a plane will make the highest rate of turn, the smallest radius of turn and will not be overstressed.

*Lets take the SR-71 through the equations

*Top speed is Mach 3.5 that’s 2314kts

*What is the rate of turn at 80° bank?

*

*2.67 degrees/sec

*So almost standard rate

*Radius of turn with a 30° bank

*

*

*

*824,162 feet or 135 NM

*This means that the pilot would have to start his turn over MWH to be able to roll out south over the pacific coast

*

*The Mighty B-19

*That means to make the smallest, tightest turn in the B-19 we have to be at 116kts, 76.86 degrees and 4.4 Gs

*Of course we have to be loaded for utility category

*If loaded for normal category 116kts, 74.74 degrees and 3.8 Gs

*Using the chart on pg 217 that puts our radius of turn at just under 350 feet

*So if you were wondering if you could turn your B-19 inside the goal posts…

*No!

*116kts? In a B-19? In a 75 degree turn? Right!

*

*Limits by aerodynamic and structure

*The turn may be limited by stall or stress related relationships.

*Because of the induced drag generated by the needed increase lift in the turn you may not have enough thrust to overcome the drag.

*In this situation, the plane may not reach its structural limits.

*Home Work Extra Credit: Draw a Vg Diagram for the Mighty B19.  Include your equations and work on the back.

*I will replace your lowest homework score with the score on the Vg Diagram.

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